sol02 - 1 University of California Berkeley Physics H7C...

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1 University of California, Berkeley Physics H7C Spring 2011 ( Yury Kolomensky ) SOLUTION TO PROBLEM SET 2 Special Relativity Maximum score: 200 points Due: February 4, 2011 Composed and formatted by P. Pebler, Yu. Kolomensky, and T.R. Beals 1. (30 points) Starting with the transformation for velocities and the definition of relativistic energy and momentum, prove that energy E and momentum pc of a given particle fol- low Lorentz transformations, and form a relativistic 4- vector. 1. Solution Consider an inertial reference frame O , in which the particle’s velocity be u . In this frame, the particle’s energy-momentum 4-vector is pc = ( γ u mc 2 , γ u mu x c, γ u mu y c, γ u mu z c ) , where γ u := (1 - u 2 /c 2 ) 1 / 2 . Now, let us consider a second in- ertial reference frame O , moving with velocity v with respect to O . Without loss of generality, we can assume that v = v ˆx , but we cannot assume that u || v –that is, we cannot assume that u and v are colinear (for details on why we can assume v lies in the x-direction, see the solution to problem 10 from problem set 1). In the frame O , our particle has some new velocity u : u = v + u || + 1 - v 2 /c 2 u 1 + v · u /c 2 = ( v + u x ) ˆx + 1 - v 2 /c 2 ( u y ˆy + u z ˆ z ) 1 + vu x /c 2 , and thus has a new energy-momentum 4-vector: p c = ( γ u mc 2 , γ u mu x c, γ u mu y c, γ u mu z c ) = γ u mc c, v + u x 1 + vu x c 2 , u y 1 - v 2 c 2 1 + vu x c 2 , u z 1 - v 2 c 2 1 + vu x c 2 , where γ u is as follows: γ u = 1 - u 2 c 2 - 1 / 2 = 1 - ( v + u x ) 2 + (1 - v 2 c 2 )( u 2 y + u 2 z ) (1 + vu x c 2 ) 2 c 2 - 1 / 2 = 1 + vu x c 2 c c 2 - v 2 - u 2 + u 2 v 2 c 2 = 1 + vu x c 2 1 - v 2 c 2 1 - u 2 c 2 = γ u γ v 1 + vu x c 2 , where γ v = 1 / 1 - v 2 c 2 . Thus, we see that the particle’s new energy-momentum 4-vector is just the Lorentz transform of the old 4-vector: p c = γ u γ v mc 2 + mu x v , γ v m ( v + u x ) c, mu y c, mu z c . It’s also worth noting that p μ p μ /c 2 = m , or, in other words, the invariant quantity preserved by Lorentz transformations of the energy-momentum 4-vector is the rest mass , m . 2. (20 points) A charged particle of charge e and momentum p is trav- eling in a uniform magnetic field B p . Show that the trajectory is a circle of radius R , and that the radius and momentum are related by the expression p = eBR (in SI units). 2. Solution
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2 To keep this problem easy, just do everything in the “lab frame”. All the usual laws of non-relativistic mechanics apply, except that we use γm in place of m . Thus, the force on a charge e moving with velocity v = p / ( γm ) in a magnetic field B is just F = γm a = e p γm × B . Since B p , the acceleration is always perpendicular to the direction of motion and to the magnetic field, which means that the particle undergoes circular motion in the plane perpendicular to B . We know that for circular motion, a = v 2 /R , so we have the following relation: a = e p ( γm ) 2 B = v 2 R .
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